How do you implicitly differentiate #2=(x+2y)^2-xy-e^(3x+7y) #?

Answer 1

#y'=(3e^(3x+7y)-3y-2x)/(3x+8y-7e^(3x+7y)) #

#0=2(x+2y)(1+2y')-[xy'+y]-e^(3x+7y) (3+7y')# #0=2x+4xy'+4y+8yy'-xy'-y-3e^(3x+7y)-7y'e^(3x+7y)# #3e^(3x+7y)-3y-2x=3xy'+8yy'-7y'e^(3x+7y)# #3e^(3x+7y)-3y-2x=y'(3x+8y-7e^(3x+7y))# #(3e^(3x+7y)-3y-2x)/(3x+8y-7e^(3x+7y)) = y'#
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Answer 2

To implicitly differentiate ( 2 = (x + 2y)^2 - xy - e^{3x + 7y} ), we differentiate both sides of the equation with respect to ( x ).

First, differentiate each term using the chain rule, product rule, and power rule where necessary. Then, solve for ( \frac{dy}{dx} ).

After differentiating, the equation becomes:

( 0 = 2(x + 2y) \frac{dx}{dx} - (x + 2y) \frac{dy}{dx} - y - e^{3x + 7y}(3 + 7\frac{dy}{dx}) )

Now, solve for ( \frac{dy}{dx} ) by isolating it on one side of the equation.

Finally, simplify the expression for ( \frac{dy}{dx} ) if possible.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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